What is the range of (x^2+5)/(x^2-3) ?

The range of (x^2+5)/(x^2-3) is f(x) 1

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range of (x^2+5)/(x^2-3)

range

\frac{x ^{2} + 5}{x ^{2} - 3}

f (x) \leq - \frac{5}{3} or f (x) > 1

Interval Notation

(- \infty, - \frac{5}{3}] \cup (1, \infty)

Solution steps

Rewrite as

\frac{x ^{2} + 5}{x ^{2} - 3} = y

Multiply both sides by

x^{2} - 3

x^{2} + 5 = y (x^{2} - 3)

The range is the set of y for which the discriminant is greater or equal to zero

Discriminant

x^{2} + 5 = y (x^{2} - 3) : 12 y^{2} + 8 y - 20

12 y^{2} + 8 y - 20 \geq 0 : y \leq - \frac{5}{3} or y \geq 1

Check if the range interval endpoints are included

y = - \frac{5}{3} :

Included

y = 1 :

Excluded

Therefore the range is

f (x) \leq - \frac{5}{3} or f (x) > 1

\frac{- 3 x}{3 x - 4}

f (x) = \frac{9 x - 8}{2 - x}

(- 2, - 6), (- 7, 1)

f (x) = \frac{x}{1 + 2 x ^{2}}

f (x) = \frac{15 - 3 x}{x ^{2} - 8 x + 15}

What is the range of (x^2+5)/(x^2-3) ?
The range of (x^2+5)/(x^2-3) is f(x)<=-5/3 \lor f(x)>1